# Why the pullback bundle is a submanifold?

Is $$f^*E$$ (the pullback of a smooth bundle) an embedded submanifold of $$N\times E$$?

I know that it's well defined and at least immersed:

Let $$\pi: E \longrightarrow M$$ a smooth fiber bundle and $$f:M \longrightarrow N$$ a smooth function. The Poor's book constructs the pullback bundle as the subset $$f^*E:=\{(p,x) \in N\times E|\; f(p)=\pi(x)\}$$ with projection $$(f^*\pi)(p,x)=p$$ and "trivialization charts":

$$\Psi:(f^*\pi)^{-1}[f^{-1}[U]] \longrightarrow f^{-1}[U] \times F$$

$$\Psi(p,x)=(p,pr_2(\psi(x)))$$

With $$\psi:\pi^{-1}[U] \longrightarrow U\times F$$, a bundle chart of $$\pi:E \longrightarrow M$$

I've already proofed that there exist a (unique) smooth structure in $$f^*E$$ such that $$f^*\pi:f^*E\longrightarrow N$$ is a smooth fiber bundle with trivialization charts $$(f^{-1}[U],\Psi)$$. By calculus in these charts, the inclusion $$i:f^*E \longrightarrow N\times E$$ is an immersion, so $$f^*E$$ is an immersed submanifold of $$N\times E$$.

Is it an embedded submanifold? Why? Is there an easy way to show that the inclusion is an embedding?

Thanks

I found a post with the same question, the answer is interesting because it involves less structure, the minmum to apply transversality, but I think it has to exist a more easy way in the case. The Pullback Bundle is an Embedded Submanifold of its Parent Space

• No, you have things backwards. If $E$ is a bundle on $M$, then you need a smooth function $f\colon N\to M$ (not the other way around), and then $f^*E$ will be the bundle on $N$ whose fiber over $p$ is the fiber of $E$ over $f(p)$. – Ted Shifrin Dec 10 '19 at 22:39

Let $$\phi : M_1 \to M_2$$ be a smooth map between smooth manifolds $$M_1, M_2$$. It is called a (smooth) embedding iff $$\phi(M_1)$$ is a smooth submanifold of $$M_2$$ and $$\phi : M_1 \to \phi(M_1)$$ is a diffeomorphism. It is called an immersion if all derivatives $$T_p \phi : T_p M_1 \to T_{\phi(p)} M_2$$ are injective. Each immersions is locally an embedding (which means that each $$p \in M_1$$ has an open neighborhood $$U$$ such that $$\phi\mid_U$$ is an embedding). Immersions are in general no embeddings (even if they are injective). However, if an immersion is a topological embedding (which means that $$\phi : M_1 \to \phi(M_1)$$ is a homeomorphism), then $$\phi$$ is an embedding. See for example https://www.math.lsu.edu/~lawson/Chapter6.pdf.
Starting with the subspace $$M_1 = f^*E$$ of $$M_2 = N \times E$$, you have shown that the space $$M_1$$ can be endowed with a smooth structure which is uniquely determined by suitable requirements. You have moreover shown that the inclusion $$i : M_1 \to M_2$$ is an immersion. But $$i$$ is a topogical embedding by definition. Thus $$f^*E = i(f^*E)$$ is a smooth submanifold of $$N \times E$$.
• Thanks, now I understand. For me $M_1$ was a subset, I defined its topology as the unique such that $\Psi$ were bundle charts. I see that the functions $\Psi$ and $\Psi^{-1}$ are continous putting the subspace topology in $M_1$. So the previuos topology I've found is the subspace topology and the inclusion is a topological (and smooth) embedding. – alexpglez98 Dec 15 '19 at 18:22